A275288 Least k such that there exists a sequence b_1 < b_2 < ... < b_t = k that includes n and has a reciprocal sum of 1.

1, 6, 6, 12, 20, 6, 28, 24, 18, 15, 33, 12, 65, 28, 15, 48, 85, 18, 76, 20, 28, 33, 115, 24, 100, 52, 54, 28, 145, 30, 217, 96, 33, 85, 35, 36, 296, 95, 52, 40, 246, 42, 301, 55, 45, 138, 329, 48, 196, 75, 102, 52, 371, 54, 55, 56, 76, 174, 531, 60, 305, 155

Offset: 1

Author

Peter Kagey, Aug 18 2016

Keywords

• nonn

Comments

From Robert Price, Jan 04 2017: (Start)

a(11) = 33 [2,3,11,22,33]

65 >= a(13) > 26 [2,3,13,26,52,60,65]; no better solution with fewer than 15 terms.

48 >= a(16) > 32 [2,3,16,18,36,48]; no better solution with fewer than 24 terms.

85 >= a(17) > 34 [2,3,15,17,34,85]; no better solution with fewer than 12 terms.

76 >= a(19) > 19 [2,3,12,19,57,76]; no better solution with fewer than 12 terms.

a(20) = 20 [2,4,5,20]

a(21) = 28 [2,4,8,21,24,28]

a(22) = 33 [2,4,11,20,22,30,33]

115 >= a(23) > 23 [2,3,10,23,69,115]; no better solution with fewer than 11 terms.

a(24) = 24 [2,3,8,24]

100 >= a(25) > 25 [2,3,10,25,60,100]; no better solution with fewer than 11 terms.

52 >= a(26) > 26 [2,3,12,26,39,52]; no better solution with fewer than 16 terms.

54 >= a(27) > 27 [2,3,12,27,36,54]; no better solution with fewer than 9 terms.

a(28) = 28 [2,3,12,21,28]

145 >= a(29) > 29 [2,4,5,29,116,145]; no better solution with fewer than 9 terms.

a(30) = 30 [2,3,12,20,30]

217 >= a(31) > 31 [2,3,9,31,93,126,217]; no better solution with fewer than 9 terms.

96 >= a(32) > 32 [2,3,9,32,72,96]; no better solution with fewer than 11 terms.

a(33) = 33 [2,3,11,22,33]

85 >= a(34) > 34 [2,3,17,20,34,60,85]; no better solution with fewer than 9 terms.

a(35) = 35 [2,3,14,15,35]

a(36) = 36 [2,3,12,18,36]

296 >= a(37) > 37 [2,3,8,37,148,222,296]; no better solution with fewer than 8 terms.

95 >= a(38) > 38 [2,4,5,38,76,95]; no better solution with fewer than 11 terms.

52 >= a(39) > 39 [2,4,6,26,39,52]; no better solution with fewer than 15 terms.

a(40) = 40 [2,3,10,24,40]

246 >= a(41) > 41 [2,3,8,41,120,205,246]; no better solution with fewer than 9 terms.

a(42) = 42 [2,3,7,42]

192 >= a(64) [2,3,8,48,64,192]; no better solution with fewer than 9 terms.

162 >= a(81) [2,3,8,72,81,108,162]; no better solution with fewer than 9 terms.

384 >= a(128) [2,3,7,96,128,336,384]; no better solution with fewer than 8 terms.

486 >= a(243) [2,3,7,81,243,336,432,486]; no better solution with fewer than 9 terms.

a(216) = 216 [2,3,8,27,216]

196 >= a(49) [2,3,8,49,98,168,196]; no better solution with fewer than 8 terms.

a(100) = 100 [2,4,5,25,100]

363 >= a(121) [2,3,7,121,176,242,336,363]; no better solution with fewer than 8 terms.

a(144) = 144 [2,3,7,112,126,144]

a(196) = 196 [2 ,3,7,84,147,196]

a(225) = 225 [2,3,9,25,90,225]

a(500) = 500 [2,4,5,25,125,500]

It appears that in most cases a(n) is a small multiple of n. For example: a(8)=3*8, a(11)=3*11, a(35)=1*35.

If not a small multiple of n, then a small rational times n. For example: a(10)=3/2*10, a(21)=4/3*21, a(22)=3/2*22.

Conjectures:

a(2^n) = 3*n

a(3^n) = 2*n

a(5^n) = 4*n

a(6^n) = n

a(7^n) = 4*n

(End)

From Peter Kagey, Jul 20 2017: (Start)

a(n) = n if and only if n is in A092671.

Every term in this sequence is in A092671.

a(a(n)) = a(n); that is, this sequence is idempotent.

(End)

From Jon E. Schoenfield, Feb 15 2020: (Start)

For any n > 1, let P be the largest divisor of n that is either a prime (p) or prime power (p^e, where e > 1). Then a(n) >= m*P where m is the smallest integer such that p divides the numerator of the sum of some subset of the set of unit fractions {1/1, 1/2, 1/3, ..., 1/m} that includes 1/(n/P).

Conjecture (confirmed for all n <= 40000): for all n > 1, the lower bound given above is tight, i.e., a(n) = m*P where m and P are as defined above. (See Example section.) (End)

Examples

			a(1)  =  1 via [1]
a(2)  =  6 via [2, 3, 6]
a(3)  =  6 via [2, 3, 6]
a(4)  = 12 via [2, 4, 6, 12]
a(5)  = 20 via [2, 4, 5, 20]
a(6)  =  6 via [2, 3, 6]
a(7)  = 28 via [2, 4, 7, 14, 28]
a(8)  = 24 via [2, 3, 8, 24]
a(9)  = 18 via [2, 3, 9, 18]
a(10) = 15 via [2, 3, 10, 15]
a(11) > 30
a(12) = 12 via [2, 4, 6, 12]
a(13) > 30
a(14) = 28 via [2, 4, 7, 14, 28]
a(15) = 15 via [2, 3, 10, 15]
a(16) > 30
a(17) > 30
a(18) = 18 via [2, 3, 9, 18]
From _Jon E. Schoenfield_, Feb 15 2020: (Start)
For n=31, the largest prime or prime power divisor of n is P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6} has no subset sum that includes 1/(n/P) = 1/1 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7} does have such a subset sum, namely, 1/1 + 1/3 + 1/7 = 31/21, so a(31) >= 7*31 = 217. In fact, the numbers 1*31=31, 3*31=93, and 7*31=217 are elements of many sets of integers that include n=31, include no element > 217, and have a reciprocal sum of 1 (one such set is {2,3,12,28,31,93,217}), so a(31)=217.
For n=62, the largest prime or prime power divisor of n is again P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4} has no subset sum that includes 1/(n/P) = 1/2 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5} does have such a subset sum, namely, 1/2 + 1/3 + 1/5 = 31/30, so a(62) >= 5*31 = 155. In fact, the numbers 2*31=62, 3*31=93, and 5*31=155 are elements of many sets of integers that include n=62, include no element > 155, and have a reciprocal sum of 1 (one such set is {2,3,12,20,62,93,155}), so a(62)=155.
(End)
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006255, A092671, A272036, A272083.

Programs

Mathematica

Table[SelectFirst[Range@ 20, MemberQ[Map[Total, 1/DeleteCases[Rest@ Subsets[Range@ #, #], w_ /; FreeQ[w, n]]], 1] &] /. k_ /; MissingQ@ k -> 0, {n, 12}] (* Michael De Vlieger, Aug 18 2016, Version 10.2, values of a(n) > 20 appear as 0 *)

Extensions

a(11)-a(12) from Robert Price, Jan 07 2017

a(13)-a(58) from David A. Corneth, Jul 20 2017

a(59)-a(62) from Jon E. Schoenfield, Feb 15 2020